Tính giới hạn: $\lim_{x \to 0} $ $\frac{5sin^5x}{(e^x -1)^5}$

Đáp án:

$5.$

Giải thích các bước giải:

$\displaystyle\lim_{x \to 0} \dfrac{5\sin^5x}{(e^x-1)^5}\\ =5\displaystyle\lim_{x \to 0} \left(\dfrac{\sin x}{e^x-1}\right)^5\\ =5\left(\displaystyle\lim_{x \to 0} \dfrac{\sin x}{e^x-1}\right)^5\\ =5\left(\displaystyle\lim_{x \to 0} \dfrac{\sin x'}{(e^x-1)'}\right)^5\\ =5\left(\displaystyle\lim_{x \to 0} \dfrac{\cos x}{e^x}\right)^5\\ =5\left(\dfrac{\cos 0}{e^0}\right)^5\\ =5.1^5\\ =5.$

Đáp án:

$\:\:\:\lim \:_{x\:\to \:\:0}\:\:\dfrac{5sin^5x}{\left(e^x\:-1\right)^5}$ $=5$

Giải thích các bước giải:

$\:\:\:\lim \:_{x\:\to \:\:0}\:\:\dfrac{5sin^5x}{\left(e^x\:-1\right)^5}$

$=5\:.\:\lim \:\:\:_{x\to \:\:0}\dfrac{\sin \:\:\:\:^5\left(x\right)}{\left(e^x-1\right)^5}$

$=5\:.\:\lim \:_{x\to 0}\left(\dfrac{\sin \:\:\left(x\right)}{e^x-1}\right)^5$

$=5\:\left(\lim \:_{x\to 0}\dfrac{\sin \:\:\left(x\right)}{e^x-1}\right)^5$

$=5\:\left(\lim \:_{x\to 0}\dfrac{\cos \:\:\left(x\right)}{e^x}\right)^5$

$=5\left(\dfrac{\cos \:\:\left(0\right)}{e^0}\right)^5$

$=5$